Math 116 – CABLE TV
Chapter 19– Distribution of Sample Proportions
1) According to the Federal Communications Commission, in 2002, 70% of all U.S. households had cable television.
a) Give the shape, mean and standard deviation of the distribution of sample proportions for samples of size 300.
Recall: Population: all US households in 2002, and the success attribute is having cable TV
n = 300, p = .70
Since both are larger than 10, the distribution of sample proportions is approximately normally distributed
- the mean of equals the population proportion: = p = 0.70
- the standard deviation is
b) What is the probability that in a sample of 300 households at least 230 (i.e. 76.7%) of them have cable television? Does this result contradict the Federal Communications Commission information? Explain.
The sample proportion is
0.0057 = 57 / 10,000,
If 70% of US households have cable TV, observing 230 of them (i.e. 76.7%) with cable TV in a random sample of 300 is very unusual. The chances are 57 in 10-thousand cases.
This unusually high result (p-hat = 0.767 = 76.7%) may suggest that probably, the percentage of US households having cable TV is higher than 70%.
The 70% figure corresponds to 2002. This sample may have been selected at a later year in which more than 70% of US households had cable TV.
Math 116 – CABLE TV
Chapter 19 – Confidence Intervals about a Population Proportion
2) In 2005 a study was done in order to estimate the percentage of U.S. households that had cable television. A sample of 300 households was selected and 230 of them had cable television.
a)Determine the point estimate.
b)Verify that the requirements for constructing a confidence interval about p-hat are satisfied.
c)Construct a 95% confidence interval estimate for the percentage of households that had cable TV in 2005.
d)We are __95___% confident that the percentage of US households that had cable TV in 2005 was between ___71.9% and ____81.5%
e)With __95% confidence we can say that in 2005, ___76.7_% of US households had cable TV with a margin of error of __4.8%
f)The statement “95% confident” means that, if 100 samples of size __300___ were taken, about _95_ of the intervals will contain the parameter p and about __5_ will not.
g)In 2002, the Federal Communications Commission reported that 70% of all U.S. households had cable television. Does the interval suggest that the percentage has changed? Explain.
The interval is completely above 70%. With 95% confidence we can say that in 2005 the percentage of US households with cable TV was higher than in 2002.
h)Sample size - How many more households should be included in the sample to be 95% confident that a point estimate p-hat will be within 1% of p? Assume a p-hat of .767
6866 – 300 = 6566Select 6566 US households more
i)Sample size - If no preliminary sample is taken to estimate p, how large a sample is necessary to be 95% confident that the point estimate p-hat will be within a distance of 0.01 from p?
(Much larger sample, more expensive study)
We’ll need to select a sample of 9604 US households, determine the proportion that has cable TV, and then construct the confidence interval about this new estimate.
This new point estimate will be within 1% of the true population proportion.
Math 116 – CABLE TV
Chapter 19 – Testing a Proportion p
3) According to the Federal Communications Commission, in 2002, 70% of all U.S. households had cable television. In 2005 a sample of 300 households was selected and 230 of them had cable television. Test the claim that in 2005 the percent of US households with cable TV was higher than in 2002. Use a 2.5% significance level.
Population: U.S. households
Success attribute: having cable TV
- Set both hypothesis
p = 0.70
p > 0.70
- Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.
You do this. The point estimate is p-hat = 230/300 = .767
****You should be wondering: Is p-hat = .767 higher than p = 0.7 by chance, or is it significantly higher? The p-value found below will help you in answering this.
- Use a feature of the calculator to test the hypothesis. Indicate the feature used and the results:
Use 1-Prop-Z-Test and get the same values shown in the work below
Test statistic =
p-value = = P(p-hat > .767) = P(z > 2.53) = 1 – .9943 = .0057 < .025 (significance level)
***How likely is it observing a p-hat = .767 or more when you select a sample of size 300 from a population that has a proportion of successes of 0.7?
very unlikely
*** Is p-hat = .767 higher than p = 0.7 by chance, or is it significantly higher? SIGNIFICANTLY HIGHER
*****Such a p-hat would be a more likely event if we select a sample from a population in which p > .70. This is why we conclude...... (see ***** below)
- What is the initial conclusion with respect to Ho and H1?
Reject Ho and support H1
- Write the conclusion using words from the problem
****** At the 5% significance level, we support the claim that in 2005, the proportion of US households with cable TV was higher than .7
Math 116 – CABLE TV
Chapter 20 – Confidence Intervals and Hypothesis Testing about Two Population Proportions
4) In 2005 a study was done to estimate the proportion of US households with cable TV. In a sample of 300 US households, it was found that 230 of them had cable TV. Erica conducted a second study in Du Page County and found that 734 out of 1000 households had cable TV.
Population 1 – US households
Population 2: Du Page County households
Success attribute: having cable TV
a) Test her claim that the two proportions are different. Use a 5% significance level.
- Set both hypothesis
- Sketch graph, shade rejection region, label, and indicate possible locations of the point estimate in the graph.
You do this
Point estimate =
****You should be wondering: Are the two sample proportions different by chance, or are they significantly different? The p-value found below will help you in answering this.
- Use a feature of the calculator to test the hypothesis. Indicate the feature used and the results:
Test statistic = 1.13
p-value = 2*P() = 2*P(z > 1.13) = 2(1-.8708) = .257
***How likely is it observing such a difference or a more extreme one when you select samples from two populations that have the same proportions?
VERY LIKELY
*** Are the two sample proportions differentby chance, or are they significantly different? DIFFERENT BY CHANCE
- What is the initial conclusion with respect to Ho and H1?
We fail to reject Ho. We don’t have enough evidence to support H1
- Write the conclusion using words from the problem
At the 2.5% significance level, we don’t have enough evidence to support the claim that the proportion of households with cable TV in Du Page County is different from the proportion with cable TV for all US households.
b)Construct a 95% confidence interval estimate for the difference p1 – p2 and interpret the results. What is the interval suggesting?
-.0225 < < .08781
Since the interval contains zero, it’s possible for the two proportions to be equal.
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